0

我在创建警报对话框时遇到问题，该对话框显示 - 无法添加窗口 - 令牌 null 不适用于应用程序

public class Authenticator {

public static final String TAG = "Authenticator";

public static int getUserId(final String username, final String password) { 

    int retVal = 0;

    final QuickTexterApplication qta = QuickTexterApplication.getQuickTexterApplication();
    final Handler handler =  new Handler();

    Thread thread = new Thread(new Runnable() {
        public void run() {                
            Runnable displayGUIRun = new Runnable() {
                public void run() {
                    int userId = 0;
                    HttpURLConnection urlConnection = null;
                    String urlAuthenticator = qta.getResources().getString(R.string.urlAuthenticator);                      
                    try{
                        URL url = new URL(urlAuthenticator);
                        urlConnection = (HttpURLConnection) url.openConnection();

                        urlConnection.setRequestMethod("POST");         
                        urlConnection.setDoInput(true);
                        urlConnection.setDoOutput(true);

                        DataOutputStream wr = new DataOutputStream(urlConnection.getOutputStream());

                        wr.writeBytes("username=" + username + "&");
                        wr.writeBytes("password=" + password);
                        wr.flush();
                        wr.close();

                        urlConnection.connect();

                        if (urlConnection.getResponseCode() == HttpURLConnection.HTTP_OK){
                            Log.d(TAG,"HTTP OK");
                            InputStream inStream = urlConnection.getInputStream();
                            BufferedReader in = new BufferedReader(new InputStreamReader(inStream));

                            String inLine = in.readLine();                  
                            in.close();

                            Log.d(TAG,"inLine: " + inLine);
                            userId = Integer.parseInt(inLine);
                        }
                        else {
                            Log.d(TAG,"HTTP NOT OK");
                        }

                        String alertMsg = "Unable to establish connection to server";
                        switch(userId){
                            case -1 :
                                alertMsg = "You entered an invalid username or password";
                            case 0  : // This is where the exception occurs
                                AlertDialog.Builder alertBuilder = new AlertDialog.Builder(qta.getApplicationContext());
                                Log.d(TAG, alertMsg);
                                alertBuilder.setMessage(alertMsg)
                                            .setNeutralButton("Ok", new DialogInterface.OnClickListener(){
                                                @Override
                                                public void onClick(DialogInterface dialog, int id) {
                                                    dialog.cancel();
                                                }

                                            });
                                AlertDialog alert = alertBuilder.create();
                                alert.show();
                                break;

我不能使用“this”作为上下文，因为 Authenticator 不是 Avitvity。
但随后 getApplicationContext() 也不起作用......
我想做的是从 Activity 类调用方法：getUserId() 。

4

1 回答 1

0

您必须在 Authenticator 类中创建构造函数，该构造函数将为您的类提供上下文。并通过调用此构造函数将上下文传递给此类。

  public class Authenticator {
       Context myContext;

   public Authenticator(YourActivity activity)
   {
    // TODO Auto-generated constructor stub
    this.myContext = activity;          
    }
  }

通过这种方式，您可以获得此类的上下文。让我知道它是否有效。:)

于 2012-07-27T13:09:17.887 回答